Question Guide – Turning Effect of Forces and Density

Question:
A uniform rod has a wooden section and a solid rubber handle, as shown in the diagram below.
The length of the handle is l and the length of the wooden section is 4.00 l. The rod balances at a distance of 2.10 l from the rubber end.
Uniform Rod with Pivot
Assume that the density of wood is approximately 1.01 g/cm3.
What is the density of rubber?
Turning Effect of Forces and Density Multiple Choice

Solution:

The Centre of Gravity (CG) of a uniform body is at the centre of the body.

Thus, the CG of the wood section is at
Equation to find CG of Wood

Therefore, the CG of the wood section from the pivot,
Equation to find CG of Wood from Pivot

The CG of the rubber section is at
Equation to find CG of Rubber

Therefore, the CG of the rubber section from the pivot,
Equation to find CG of Rubber from Pivot

Since the rod is at equilibrium, taking moment (M) about the pivot,
Taking moment at pivot

Since the weight of a body is the product of its mass (m) and the gravitational field strength (g),
Weight is is the product of its mass and the gravitational field strength

And since the density () of a body is its mass per unit volume (V),
Density equals to mass divided by Volume

By substituting equation (3) into (2), we get
Substitute mass with density times volume

And since volume is the product of cross-sectional area (A) and length (L),
Volume equals to Area times length
By substituting equation (5) into (4), we get
Substitute Volume with Area times Length

By substituting equation (6) into (1), we get
Final calculation and equations for density

Thus, Answer is B.

This question is related to the topic “Turning Effect of Forces” and “Mass, Weight and Density” in Physics tuition for O Level.

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Question Guide – Turning Effect of Forces and Density ultima modifica: 2017-10-24T07:49:13+00:00 da LearningForKeeps